Examples

Chapter 12 Class 11 Limits and Derivatives
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Example 20 Find the derivative of f(x) from the first principle, where f(x) is (ii) x sin x Given f (x) = x sin x We need to find Derivative of f(x) We know that fβ(x) = limβ¬(hβ0) πβ‘γ(π₯ + β) β π(π₯)γ/β Here, f (x) = x sin x So, f (x + h) = (x + h) sin (x + h) Putting values fβ(x) =limβ¬(hβ0) ((π₯ + β) sinβ‘γ (π₯ + β) β π₯ sinβ‘γπ₯ γ γ)/β Using sin (A + B) = sin A cos B + cos A sin B = limβ¬(hβ0)β‘γ((π₯ + β)(sinβ‘γπ₯ cosβ‘γβ +γ cos π₯γβ‘sinβ‘γβ γ )γ β π₯ sinβ‘π₯ γ)/βγ = limβ¬(hβ0)β‘γ(π₯(sinβ‘γπ₯ cosβ‘γβ +γ cosγβ‘γπ₯ sinβ‘γβ) + β (sinβ‘γπ₯ cosβ‘γβ +γ cosγβ‘γπ₯ sinβ‘γβ) β π₯ sinβ‘π₯ γ γ γ γ γ γ γ γ)/βγ = limβ¬(hβ0)β‘γ(π₯ sinβ‘γπ₯ cosβ‘γβ + π₯ cosβ‘γπ₯ sinβ‘γβ + β γ(sinγβ‘γπ₯ cosβ‘γβ + cosβ‘γπ₯ sinβ‘γβ) β π₯ sinβ‘π₯ γ γ γ γ γ γ γ γ)/βγ = limβ¬(hβ0)β‘γ(π₯π ππ π₯ cosβ‘γβ β π₯ sinβ‘γπ₯ + π₯ cosβ‘γπ₯ sinβ‘γβ + β(sinβ‘γπ₯ cosβ‘γβ + cosβ‘γπ₯ sinβ‘β γ γ)γ γ γ γ γ)/βγ = limβ¬(hβ0)β‘γ(π₯π ππ π₯ γ(cosγβ‘γβ β 1)+ π₯ cosβ‘γπ₯ sinβ‘γβ + β(sinβ‘γπ₯ cosβ‘γβ + cosβ‘γπ₯ sinβ‘β γ γ)γ γ γ γ)/βγ = limβ¬(hβ0)β‘((π₯ sinβ‘γπ₯ (cosβ‘γβ β1)γ γ)/β+(π₯ cosβ‘γπ₯ sinβ‘β γ)/β+(β (sinβ‘γπ₯ cosβ‘γβ +cosβ‘γπ₯ sinβ‘γβ)γ γ γ γ)/β) = limβ¬(hβ0)β‘γγx sinγβ‘γπ₯ (cosβ‘γβ β1)γ γ/β+limβ¬(hβ0) (π₯ cosβ‘γπ₯ sinβ‘β γ )/h+limβ¬(hβ0) (sinβ‘γπ₯ cosβ‘γβ +cosβ‘γπ₯ sinβ‘γβ )γ γ γ γ γ = β x sin x (π₯π’π¦)β¬(π‘βπ) ((πβγ πππγβ‘γπ)γ)/π+π₯ cosβ‘γπ₯ (π₯π’π¦)β¬(π‘βπ) πππβ‘π/π‘+γ limβ¬(hβ0) (sinβ‘γπ₯ cosβ‘γβ +cosβ‘γπ₯ sinβ‘γβ )γ γ γ γ = β x sin x (0) + x cos x (1) + ( sin x cos 0 + cos x sin 0) = 0 + x cos x + sin Γ 1 + cos x Γ 0 = 0 + x cos x + sin x + 0 = x cos x + sin x Hence fβ (x) = x cos x + sin x